Free CAS MAS-I Statistics Practice Questions (2026)

Sample Questions

Question 1 Easy

An actuary records five independent claim severities (in thousands):

8, \; 12, \; 15, \; 18, \; 22

Calculate the unbiased sample variance.

Solution

D is correct. First compute the sample mean:

\bar{X} = \frac{8 + 12 + 15 + 18 + 22}{5} = \frac{75}{5} = 15.

The deviations from the mean are

-7, -3, 0, 3, 7

, with squared deviations

49, 9, 0, 9, 49

, summing to

116

. The unbiased sample variance uses the divisor

n - 1 = 4

S^2 = \frac{\sum (X_i - \bar{X})^2}{n - 1} = \frac{116}{4} = 29.

The value 29 falls in the interval

[25, 30)

, so the answer is D: at least 25, but less than 30.

Question 2 Medium

Let

X_1, X_2, X_3, X_4, X_5

be a random sample of five independent observations from a continuous Uniform

(0, 10)

distribution. Calculate the probability that the sample maximum

X_{(5)}

exceeds 8.

Solution

D is correct. The maximum

X_{(5)}

is at most 8 if and only if every observation is at most 8. By independence,

P(X_{(5)} \leq 8) = \prod_{i=1}^{5} P(X_i \leq 8) = [F_X(8)]^5 = (8/10)^5 = 0.8^5 = 0.32768.

Therefore

P(X_{(5)} > 8) = 1 - 0.32768 = 0.67232

. This value falls in the interval

[0.65, 0.80)

Question 3 Hard

An actuary observes six iid claim amounts drawn from an exponential distribution with mean

\theta = 250

. Let

X_{(1)} \leq X_{(2)} \leq \cdots \leq X_{(6)}

denote the order statistics.

Calculate

P(X_{(3)} > 200).

Solution

B is correct.

Step 1: Translate the event in terms of how many observations fall on each side of 200.

X_{(3)} > 200

means the third-smallest of the six exceeds 200, equivalently at most two of the six observations are

\leq 200.

Step 2: Define a binomial counter. Let

Y = \#\{i : X_i \leq 200\}

. Since the

X_i

are iid,

Y \sim \text{Binomial}(6, p)

with

p = P(X \leq 200).

Step 3: Compute

p

using the exponential cdf.

p = 1 - e^{-200/250} = 1 - e^{-0.8} \approx 1 - 0.4493 = 0.5507,\quad 1 - p \approx 0.4493.

Step 4: Sum the binomial probabilities for

Y \leq 2

P(Y = 0) = (0.4493)^6 \approx 0.00823

P(Y = 1) = 6(0.5507)(0.4493)^5 \approx 6 \cdot 0.5507 \cdot 0.01831 \approx 0.06051

P(Y = 2) = 15(0.5507)^2(0.4493)^4 \approx 15 \cdot 0.30327 \cdot 0.04075 \approx 0.18539.

Step 5: Add.

P(X_{(3)} > 200) = P(Y \leq 2) \approx 0.00823 + 0.06051 + 0.18539 \approx 0.254.

The single-observation tail

P(X > 200) = e^{-0.8} \approx 0.449

is a tempting shortcut but ignores the order-statistic structure entirely; it answers a different question.

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